Find Dy Dx : Tan(xy)= X. Do Not Simplify The Result.
Understanding how to differentiate implicit functions is a fundamental skill in calculus, especially when dealing with equations where the dependent and independent variables are intertwined in a non-explicit manner. The problem at hand involves the implicit relation \(\tan(xy) = x\), and the goal is to find the derivative \(\frac{dy}{dx}\) without simplifying the resulting expression. This type of problem not only tests your proficiency with implicit differentiation but also reinforces concepts like the chain rule, product rule, and the derivatives of inverse functions.
In this article, we will explore the systematic approach to differentiate the equation \(\tan(xy) = x\). We will discuss the underlying principles, step-by-step differentiation, and interpret the implications of the derivative. The methodology emphasizes clarity and precision, ensuring that each step is justified and that the final expression retains all its complexity without simplification.
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Understanding the Problem
Implicit Functions and Their Differentiation
Implicit functions are those where the dependent variable \(y\) cannot be explicitly expressed as a function of \(x\) in a straightforward manner. Instead, the relationship is given as an equation involving both \(x\) and \(y\). Differentiating such functions requires implicit differentiation, which involves differentiating both sides of the equation with respect to \(x\), treating \(y\) as a function of \(x\).
Key points to remember:
- When differentiating terms involving \(y\), multiply by \(\frac{dy}{dx}\) (the chain rule).
- The derivative of \(\tan(u)\) with respect to \(u\) is \(\sec^2(u)\).
- The derivative of \(xy\) with respect to \(x\) involves the product rule.
The Given Equation
The equation is:
\[
\tan(xy) = x
\]
Our goal is to find \(\frac{dy}{dx}\).
This involves:
- Differentiating \(\tan(xy)\) with respect to \(x\),
- Differentiating \(x\) with respect to \(x\),
- Applying the chain rule and the product rule appropriately.
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Step-by-Step Differentiation Process
Step 1: Differentiate Both Sides
Start by differentiating the entire equation with respect to \(x\):
\[
\frac{d}{dx}[\tan(xy)] = \frac{d}{dx}[x]
\]
This yields:
\[
\frac{d}{dx}[\tan(xy)] = 1
\]
Step 2: Apply Chain Rule to \(\tan(xy)\)
Recall:
\[
\frac{d}{dx}[\tan(u)] = \sec^2(u) \cdot \frac{du}{dx}
\]
Here, \(u = xy\). Therefore:
\[
\frac{d}{dx}[\tan(xy)] = \sec^2(xy) \cdot \frac{d}{dx}[xy]
\]
---
Step 3: Differentiate \(xy\) Using Product Rule
Since \(xy\) is a product of \(x\) and \(y\):
\[
\frac{d}{dx}[xy] = y + x \frac{dy}{dx}
\]
- Derivative of \(x\) is 1,
- Derivative of \(y\) with respect to \(x\) is \(\frac{dy}{dx}\).
Putting it all together:
\[
\frac{d}{dx}[\tan(xy)] = \sec^2(xy) \cdot (y + x \frac{dy}{dx})
\]
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Step 4: Differentiate the Right Side
The derivative of \(x\) with respect to \(x\) is:
\[
\frac{d}{dx}[x] = 1
\]
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Step 5: Write the Differentiated Equation
Combining the results:
\[
\sec^2(xy) \cdot (y + x \frac{dy}{dx}) = 1
\]
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Solving for \(\frac{dy}{dx}\)
Step 6: Isolate \(\frac{dy}{dx}\)
Distribute \(\sec^2(xy)\):
\[
\sec^2(xy) \cdot y + \sec^2(xy) \cdot x \frac{dy}{dx} = 1
\]
Bring all terms involving \(\frac{dy}{dx}\) to one side:
\[
\sec^2(xy) \cdot x \frac{dy}{dx} = 1 - \sec^2(xy) \cdot y
\]
Now, solve for \(\frac{dy}{dx}\):
\[
\frac{dy}{dx} = \frac{1 - \sec^2(xy) \cdot y}{\sec^2(xy) \cdot x}
\]
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Step 7: Final Expression (Without Simplification)
The derivative \(\frac{dy}{dx}\) is:
\[
\boxed{
\frac{dy}{dx} = \frac{1 - y \sec^2(xy)}{x \sec^2(xy)}
}
\]
This expression preserves all the components and respects the instruction not to simplify further.
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Discussion and Interpretation
Components of the Derivative
- The numerator, \(1 - y \sec^2(xy)\), combines the constant term with a product involving \(y\) and the secant squared of the product \(xy\).
- The denominator, \(x \sec^2(xy)\), involves the independent variable \(x\) and the same secant squared term.
Implications and Usage
- The expression can be used to analyze the slope of the tangent line to the curve defined by \(\tan(xy) = x\) at any point where the function is differentiable.
- It can be employed in further calculations, such as finding the second derivative or analyzing the behavior of the function.
Special Cases and Considerations
- When \(x = 0\), the derivative expression involves division by zero, indicating a potential point of discontinuity or where the derivative does not exist.
- When \(\sec^2(xy) = 0\), which is impossible because \(\sec^2\) is always positive, so no concern there.
- The relationship between \(x\) and \(y\) governed by the original equation constrains the domain where the derivative is valid.
Summary and Final Remarks
The process of differentiating the implicit equation \(\tan(xy) = x\) demonstrates the application of several core calculus principles:
- Implicit differentiation,
- Chain rule,
- Product rule,
- Handling composite functions involving trigonometric and algebraic components.
The final derivative, expressed as:
\[
\boxed{
\frac{dy}{dx} = \frac{1 - y \sec^2(xy)}{x \sec^2(xy)}
}
\]
remains unsimplified, fulfilling the problem's requirement. This form provides a comprehensive view of the derivative's structure and is ready for further analysis or application.
Mastering such differentiation techniques enhances problem-solving skills in calculus, especially in dealing with complex relationships between variables. It emphasizes the importance of meticulous differentiation and understanding the behavior of implicit functions in mathematical analysis.
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References:
- Stewart, James. Calculus: Early Transcendentals. 8th Edition.
- Thomas, George B., and Ross L. Finney. Calculus and Analytic Geometry. 9th Edition.
- Online calculus resources and tutorials on implicit differentiation and trigonometric derivatives.
